Special Lines, xy = k and Inequalities

The equation $x=-3$ says nothing about $y$, so $y$ can be anything, but $x$ must be $-3$: the points $(-3,-2)$, $(-3,0)$, $(-3,5)$ all work. Together they form a vertical line crossing the $x$-axis at $(-3,0)$. Its run between any two points is $0$, so its slope is undefined. The equation $y=4$ fixes $y$ and lets $x$ vary, giving a horizontal line through $(0,4)$ with slope $0$. Two special cases are worth remembering: $x=0$ is the $y$-axis itself, and $y=0$ is the $x$-axis.

The points on $x+y=10$ are pairs of numbers that add to 10, like 7 right answers and 3 wrong answers on a 10-question quiz: $(0,10)$, $(3,7)$, $(10,0)$. Solving for $y$ gives $y=-x+10$, so every line $x+y=k$ has slope $-1$ and crosses both axes at $k$. The points on $x-y=3$ are pairs whose difference is 3, like $(5,2)$ and $(3,0)$. Solving for $y$: subtract $x$ from both sides to get $-y=-x+3$, then multiply by $-1$ to get $y=x-3$. So every line $x-y=k$ has slope $1$ and $y$-intercept $-k$. Note the sign: the $y$-intercept of $x-y=3$ is $-3$, not $3$.

An equation like $2x+3y=6$ is in standard form, $ax+by=k$. The fastest way to graph it uses the two intercepts. On the $x$-axis $y=0$, so $2x=6$ and the $x$-intercept is $(3,0)$. On the $y$-axis $x=0$, so $3y=6$ and the $y$-intercept is $(0,2)$. Plot both points and draw the line through them. If you want the slope, rearrange: $3y=-2x+6$, so $y=-\frac{2}{3}x+2$. This shortcut needs two different intercepts. If $k=0$ (as in $2x+3y=0$), both intercepts are the origin, so find a second point by choosing another $x$. If $a=0$ the line is horizontal, and if $b=0$ it is vertical.

6.3 Graph Using Intercepts (Grade 9 Academic MPM1D)

The points on $xy=6$ are pairs that multiply to 6: $(1,6)$, $(2,3)$, $(3,2)$, $(6,1)$, and also $(1.5,4)$ and $(4,1.5)$. Solving for $y$ gives $y=\frac{6}{x}$. When $x$ doubles, $y$ halves, so equal steps in $x$ do not give equal changes in $y$: for $x=1,2,3$ the $y$-values $6,3,2$ have first differences $-3$ and $-1$. The relation is non-linear, and its graph is a curve. The value $x=0$ is impossible, because $0\times y=0$, never 6, so the curve never touches the $y$-axis; in the same way it never touches the $x$-axis. Negative pairs such as $(-2,-3)$ also multiply to 6, which gives a second, separate branch. (When $k$ is negative, as in $xy=-6$, the two branches sit in the other two quadrants.) In a context like a rectangle's length and width, only the positive branch makes sense.

An inequality such as $2x+3y\le 6$ is true for a whole region of points, not just a line. To graph it: (1) graph the boundary, the line you get by replacing the inequality sign with $=$; (2) draw it solid if the sign is $\le$ or $\ge$ (points on the line count) and dashed if it is $<$ or $>$ (they do not); (3) pick a test point that is not on the line, often $(0,0)$, and substitute it; (4) if the test point makes the inequality true, shade its side of the line; if false, shade the other side. For $2x+3y\le 6$, the test $(0,0)$ gives $0\le 6$, which is true, so you shade the side containing the origin.

Graphing Linear Inequalities | math 9 | jensenmath.ca